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# Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?

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Manager
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Joined: 02 Nov 2018
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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18 Mar 2019, 03:47
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Difficulty:

75% (hard)

Question Stats:

35% (02:13) correct 65% (02:01) wrong based on 23 sessions

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Let k = 2008^2 + 2^2008 . What is the units digit of $$k^2 + 2^k$$ ?

(A) 0

(B) 2

(C) 4

(D) 6

(E) 8

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Math Expert
Joined: 02 Aug 2009
Posts: 7575
Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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18 Mar 2019, 04:53
Let k = 2008^2 + 2^2008 . What is the units digit of $$k^2 + 2^k$$ ?

(A) 0

(B) 2

(C) 4

(D) 6

(E) 8

Easiest way would be to find the units digit of k...
$$k = 2008^2 + 2^{2008}$$, the units digit will be same as that of $$8^2+2^{4*502}$$, that is 4+6 = 10
thus the units digit of $$k^2 + 2^k$$ will be $$10^2+2^{4*something}$$ or 0+2^4=0+6=6

Thus D

Editing, took a wrong value of k
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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18 Mar 2019, 05:05
Hi chetan2u
this is the OE
The units digit of 2^n is 2, 4, 8, and 6 for n = 1, 2, 3, and 4,
respectively. For n > 4, the units digit of 2^n is equal to that of 2n^−4 .
Thus for every positive integer j the units digit of 2^4j is 6,
and hence 2^2008 has a units digit of 6.
The units digit of 2008^2 is 4.
Therefore the units digit of k is 0,
so the units digit of k^2 is also 0.
Because 2008 is even, both 2008^2 and 2^2008 are
multiples of 4.
Therefore k is a multiple of 4, so the units digit of 2^k is 6, and
the units digit of k^2 + 2^k is also 6
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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18 Mar 2019, 05:10
Hi chetan2u
this is the OE
The units digit of 2^n is 2, 4, 8, and 6 for n = 1, 2, 3, and 4,
respectively. For n > 4, the units digit of 2^n is equal to that of 2n^−4 .
Thus for every positive integer j the units digit of 2^4j is 6,
and hence 2^2008 has a units digit of 6.
The units digit of 2008^2 is 4.
Therefore the units digit of k is 0,
so the units digit of k^2 is also 0.
Because 2008 is even, both 2008^2 and 2^2008 are
multiples of 4.
Therefore k is a multiple of 4, so the units digit of 2^k is 6, and
the units digit of k^2 + 2^k is also 6

Right, I took the value of k wrongly as 10
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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18 Mar 2019, 05:34
2
1
chetan2u wrote:
Let k = 2008^2 + 2^2008 . What is the units digit of $$k^2 + 2^k$$ ?

(A) 0

(B) 2

(C) 4

(D) 6

(E) 8

Easiest way would be to find the units digit of k...
$$k = 2008^2 + 2^{2008}$$, the units digit will be same as that of $$8^2+2^{4*502}$$, that is 4+6 = 10
thus the units digit of $$k^2 + 2^k$$ will be $$10^2+2^{10}$$ or 0+2^2=0+4=4

Thus C

OA is wrong

We can easily conclude that the unit digit of K = 2008^2 + 2^2008 = 0

but we have to conclude the remainder K leaves when divided by 4 so as to find out the cyclicity(we can not simply conclude it as 10 as 20 having 0 as it's unit digit leaves remainder 0 as compared to 10 leaving 2)

So unit digit of K is 0 , the unit digit of $$k^2$$= 0---------(a)

2^(2008^2 + 2^2008)
=2^(2008^2) * 2^(2^2008)
=2^(value divisible by 4) * 2^(value divisible by 4)
=4*4=16
unit digit 6----- (b)

So the unit digit of $$k^2 + 2^k$$ = 0+6 =6
Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?   [#permalink] 18 Mar 2019, 05:34
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